Find the absolute maximum and absolute minimum values, if any, of the function. (If an answer does not exist, enter DNE.)
h(x) = x3 + 3x2 + 4 on [−3, 2]
Well, finding the absolute maximum and minimum values of a function can be quite challenging, just like figuring out why clowns wear such big shoes, but let's give it a shot anyway.
First, we'll start by finding the critical points of the function h(x). To do that, we can take the derivative of h(x):
h'(x) = 3x^2 + 6x
To find the critical points, we need to set h'(x) equal to zero and solve for x:
3x^2 + 6x = 0
Factoring out an x, we have:
x(3x + 6) = 0
Setting each factor equal to zero, we find that x = 0 and x = -2 are the critical points.
Next, we'll evaluate the function at the endpoints of the interval [-3, 2] and at the critical points:
h(-3) = (-3)^3 + 3(-3)^2 + 4 = -20
h(2) = 2^3 + 3(2)^2 + 4 = 20
h(0) = 0^3 + 3(0)^2 + 4 = 4
h(-2) = (-2)^3 + 3(-2)^2 + 4 = -4
Now we compare all these values. The highest value is 20, which occurs at x = 2, and the lowest value is -20, which happens at x = -3.
Therefore, the absolute maximum value of h(x) is 20, and it occurs at x = 2. As for the absolute minimum value, it is -20, and it occurs at x = -3.
So, we've found our answers, just like a clown finding laughter in the most unexpected places!
To find the absolute maximum and absolute minimum values of a function, we need to take the derivative of the function, set it equal to zero to find the critical points, and then evaluate the function at those critical points as well as the endpoints of the given interval.
Let's start by finding the critical points by taking the derivative of h(x):
h'(x) = 3x^2 + 6x
To find the critical points, we set h'(x) equal to zero and solve for x:
3x^2 + 6x = 0
3x(x + 2) = 0
Setting each factor equal to zero, we get:
3x = 0
x = 0
x + 2 = 0
x = -2
So, the critical points are x = 0 and x = -2.
Next, we evaluate the function at these critical points as well as the endpoints of the interval:
h(-3) = (-3)^3 + 3(-3)^2 + 4 = -27 + 27 + 4 = 4
h(2) = (2)^3 + 3(2)^2 + 4 = 8 + 12 + 4 = 24
h(0) = (0)^3 + 3(0)^2 + 4 = 0 + 0 + 4 = 4
h(-2) = (-2)^3 + 3(-2)^2 + 4 = -8 + 12 + 4 = 8
Now we compare these values to determine the absolute maximum and minimum:
- The function evaluates to 4 at both -3 and 0.
- The function evaluates to 24 at 2.
- The function evaluates to 8 at -2.
Therefore, the absolute maximum value of h(x) is 24, which occurs at x = 2, and the absolute minimum value is 4, which occurs at both x = -3 and x = 0.
To find the absolute maximum and absolute minimum values of the function h(x) = x^3 + 3x^2 + 4 on the interval [-3, 2], we need to follow these steps:
Step 1: Find the critical points of h(x) by finding where its derivative is equal to zero or does not exist.
Step 2: Evaluate the function at the endpoints of the interval.
Step 3: Compare the values obtained in steps 1 and 2 to determine the absolute maximum and absolute minimum values.
Step 1: Find the critical points of h(x) by finding where its derivative is equal to zero or does not exist.
To find the critical points, we need to take the derivative of h(x).
h'(x) = 3x^2 + 6x
Setting the derivative equal to zero:
3x^2 + 6x = 0
Factor out a 3x on the left side:
3x(x + 2) = 0
From here, we can see that there are two solutions:
x = 0 and x = -2
So the critical points are x = 0 and x = -2.
Step 2: Evaluate the function at the endpoints of the interval.
Evaluate h(x) at the endpoints -3 and 2:
h(-3) = (-3)^3 + 3(-3)^2 + 4 = -11
h(2) = (2)^3 + 3(2)^2 + 4 = 20
Step 3: Compare the values obtained in steps 1 and 2 to determine the absolute maximum and absolute minimum values.
We have the following points:
(-3, -11), (0, 4), (-2, -4), and (2, 20).
The absolute minimum value occurs at the point (-3, -11) since -11 is the smallest y-value.
The absolute maximum value occurs at the point (2, 20) since 20 is the largest y-value.
Therefore, the absolute minimum value of h(x) is -11, and the absolute maximum value is 20.
I will assume you meant
f(x) = x^3 + 3x^2 + 4
f ' (x) = 3x^2 + 6x = 0 for max/min
3x(x+2) = 0
x = 0 or x = -2
f(0) = 4
f(-2) = -8 + 12 + 4 = 8
end values:
f(-3) = -27 + 27 + 4 = 4
f(2) = 8 + 12 + 4 = 24
minimum value of the function is 4
maximum value is 24